Tire value of ∫1xn1+xn′1/ndx,(n∈N), is
11−n1+1xn1−1n+C
11+n1−1xn1−1H+C
−11−n1−1xn1−1n+C
−11+n1+1xn1−1n+C
Let
I=∫1xn1+xn1/ndx=∫1xn+11+1xn1/ndx⇒ I=∫1+x−n−1/nx−n−1dx⇒ l=−1n∫1+x−n−1/nd1+x−n⇒ l=−1n1+x−n−1+1−1n+1+C=11−n1+1xn1−1π+C